I have found the following two forms of the negative binomial distribution:
1) Let random variable $X$ be the number of failures before $r$ successes are obtained. Then the pmf of $X$ is given by
$$P(X = x) = {x+r-1\choose r-1}p^r(1-p)^x$$
2) Let random variable $X$ be the number of trials performed before r successes are obtained. Then the pmf of $X$ is given by
$$P(X = x) = {x-1\choose r-1}p^r(1-p)^{x-r}$$
In either case, the random variable $X$ is defined differently. Obviously, the expected value of $X$ in either case then, $\mu$ is different. In the first case, $$\mu = \frac{r(1-p)}{p}$$ and in the second case,$$\mu = \frac{r}{p}$$
Incidentally or otherwise, however, the variance $Var (X)$ is the same in either case $$Var(X) = \frac{r(1-p)}{p^2}$$
a) What is the motivation behind having alternative forms of the same probability distribution (or are they different) ?
b) Is there some intuition behind their having the same variance?
Best Answer
These two definitions are essentially equivalent.
Let $X$ be the number of failures before $r$ successes are obtained. Let $Y$ be the number of trials before $r$ successes are obtained. Suppose further that each trial is iid Bernoulli distributed.
We have that $Y=X+r$. Hence, $E(Y) = E(X)+r$, and $Var(X)=Var(Y)$.